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MaRussiya [10]
3 years ago
10

What lessons learned from 2008 cam we apply to today's current situation​

Mathematics
1 answer:
Nikolay [14]3 years ago
6 0

Answer:

Listen!

Good communication starts with listening closely and asking lots of questions. Understand what is happening. Understand what people (coworkers, customers, etc.) are feeling and confronting. Care about what they are feeling and confronting, because it is their reality. Listen before you speak, and certainly listen before you make decisions.

Be as clear as possible.

If you have listened well and understand what others are facing, you will be better able to communicate clearly because you will be answering questions before they are asked. Through the year+ of running Treasury’s HFA Initiative, we held countless webinars with large numbers of participants. This was a primary method of group communication. We spent a great deal of time preparing for those.

Anticipate and plan.

Even if you are getting pressed from every side, you need to take time to think, anticipate and plan. Anticipate plausible scenarios and plan for those scenarios. Manage toward what you want to achieve, but also plan for other plausible scenarios and be prepared. If you have leadership responsibility, others are expecting you to be prepared.

Have resolve and navigate.

Know that you will face a series of challenges and resolve that you will navigate through those challenges. If you’ve taken the time to think and plan for scenarios, you will be better prepared to navigate through whatever comes your way.

Find ways to recharge even if the pressure seems relentless.

You need to find some way to disengage, even for short periods of time, to recharge the batteries and get perspective. Meditate, pray, take a walk… whatever works for you. Energy is finite.

Keep the end in sight.

When you feel overwhelmed, know that these challenges will pass. Do your best… listen, think, plan, navigate, and push through. In the end, you will be stronger than you were when the challenge started.

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Create a quadratic trinomial with zeros equal to 4 and 5
Ilya [14]

Answer:

x^2 - 9x + 20

Step-by-step explanation:

(x - 4)(x - 5) (have to do the opposite for it to be positive)

FOIL method: X x X = x^2

-5 x X = -5x

-4 x X = -4x

x^2 - 5x - 4x + 20

-5 - 4 = -9 so...

x^2 - 9x +20 is the answer

5 0
3 years ago
Find the midpoint of the line segment defined by the points: (5, 4) and (−2, 1) (2.5, 1.5) (3.5, 2.5) (1.5, 2.5) (3.5, 1.5)
Setler79 [48]

Answer:

\boxed {\boxed {\sf (1.5 , 2.5)}}

Step-by-step explanation:

The midpoint is the point that bisects a line segment or divides it into 2 equal halves. The formula is essentially finding the average of the 2 points.

(\frac {x_1+x_2}{2}, \frac {y_1+ y_2}{2})

In this formula, (x₁, y₁) and (x₂, y₂) are the 2 endpoints of the line segment. For this problem, these are (5,4 ) and (-2, 1).

  • x₁= 5
  • y₁= 4
  • x₂= -2
  • y₂= 1

Substitute these values into the formula.

( \frac {5+ -2}{2}, \frac {4+1}{2})

Solve the numerators.

  • 5+ -2 = 5-2 = 3
  • 4+1 = 5

( \frac {3}{2}, \frac{5}{2})

Convert the fractions to decimals.

(1.5, 2.5)

The midpoint of the line segment is (1.5 , 2.5)

3 0
3 years ago
A line goes through the points (5, -1) and (7, -5). What is a possible equation to a new line that is perpendicular?
Leya [2.2K]

<em>Ok, so you are given an equation in standard form, 5x+4y=2, and a point, (7, 5). You are being asked to write the equation for a line that is parallel to the equation in standard from, and that includes the point (7, 5).  </em>

<em> </em>

<em> </em>

<em> </em>

<em>First, let's start by finding the slope of your new line. We know that it needs to have the same slope as 5x+4y=2, because parallel lines have the same slopes. To do that, we need to put the equation into slope-intercept form (y=mx+b), which means we need to isolate the "y."  </em>

<em> </em>

<em> </em>

<em> </em>

<em>  5x+4y=2 </em>

<em> </em>

<em>-5x     -5x  (subtract 5x from both sides to move it to the right side of your equation) </em>

<em> </em>

<em>4y = 2 - 5x </em>

<em> </em>

<em>/4  /4  /4 (divide all the terms by 4 to get "y" by itself) </em>

<em> </em>

<em>y = (1/2) - (5/4)x  ... I suggest leaving your slope as a fraction. </em>

<em> </em>

<em> </em>

<em> </em>

<em>Now, we know that our slope, m , is going to be -(5/4).  </em>

<em> </em>

<em> </em>

<em> </em>

<em>Next, we are going to use our slope, -(5/4), and point, (7, 5), to find the b-value (y-intercept) of your new line. Let's plug in what we know: </em>

<em> </em>

<em> </em>

<em> </em>

<em>y = 5 </em>

<em> </em>

<em>m = -(5/4) </em>

<em> </em>

<em>x = 7 </em>

<em> </em>

<em> </em>

<em> </em>

<em>y=mx+b </em>

<em> </em>

<em>5=(-5/4)(7) + b  -> I plugged in what we knew for y, m, and x.  </em>

<em> </em>

<em>5 = -(35/4) + b  -> I multiplied the numbers in the numerator (5 x 7) to get 35/4 </em>

<em> </em>

<em>20/4 = -(35/4) + b -> I converted 5 into a faction with a denominator of 4 by multiplying by (4/4) </em>

<em> </em>

<em>+(35/4) +(35/4)  -> I add (35/4) to both sides to isolate b </em>

<em> </em>

<em>55/4 = b ... or b = 13.74 </em>

<em> </em>

<em> </em>

<em> </em>

<em>Answer:  </em>

<em> </em>

<em>m = -(5/4) </em>

<em> </em>

<em>b = 13.75</em>

7 0
2 years ago
Help i need help with this question
makvit [3.9K]
The answer is the √26 and -√26. Look at the picture for explanation.

3 0
3 years ago
A quadrilateral is sometimes a square? <br><br> True or False
Vesnalui [34]
I know this ...... it is false
7 0
3 years ago
Read 2 more answers
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